Let vec{a}, vec{b} and vec{c} be vectors of e | vedclass

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(D) Let $|\vec{a}| = |\vec{b}| = |\vec{c}| = k$. We know that $\vec{a} \cdot \vec{b} = k^2 \cos \alpha$, $\vec{b} \cdot \vec{c} = k^2 \cos \beta$, and

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$-\frac{3}{2}$

Vedclass · Answer

(D) Let $|\vec{a}| = |\vec{b}| = |\vec{c}| = k$. We know that $\vec{a} \cdot \vec{b} = k^2 \cos \alpha$, $\vec{b} \cdot \vec{c} = k^2 \cos \beta$, and $\vec{c} \cdot \vec{a} = k^2 \cos \gamma$. Consider the magnitude of the sum of the vectors: $|\vec{a} + \vec{b} + \vec{c}|^2 \geq 0$. Expanding this, we get: $(\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c}) \geq 0$. $|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \geq 0$. Substituting the values: $3k^2 + 2k^2(\cos \alpha + \cos \beta + \cos \gamma) \geq 0$. Dividing by $2k^2$ (since $k > 0$): $\frac{3}{2} + (\cos \alpha + \cos \beta + \cos \gamma) \geq 0$. Therefore, $\cos \alpha + \cos \beta + \cos \gamma \geq -\frac{3}{2}$.

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be vectors of equal magnitude such that the angle between $\vec{a}$ and $\vec{b}$ is $\alpha$, $\vec{b}$ and $\vec{c}$ is $\beta$, and $\vec{c}$ and $\vec{a}$ is $\gamma$. Then the minimum value of $\cos \alpha + \cos \beta + \cos \gamma$ is

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